my answer that I have gotten is $b'c'd + a' b d'$ however, the answer given to me was b'c'd can someone tell me whether I am correct
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I calculated, the correct answer is b'c'd
Could you show the steps you used?
Amit Saxena
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(a+d)(a′b+c′d)(ac+bd)′ = (aa'b+ac'd+a'bd+c'd)(a'+c')(b'+d') – kero Nov 15 '15 at 15:54
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Yes, aa'b = 0 The product will be (a'bd+c'd)(b'+d') which will lead you to b'c'd – Amit Saxena Nov 15 '15 at 15:58
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managed to work it out, was careless! any tips on simplification since i am doing it by expanding all the terms once at a time – kero Nov 15 '15 at 16:09
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When you expand (ac'd+a'bd+c'd)(a'+c'), four of the six terms will cancel out. – Amit Saxena Nov 15 '15 at 16:13
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1530261 $F\equiv((a\lor d)\land((\lnot a\land b)\lor(\lnot c\land d))\land\lnot((a\land c)\lor(b\land d)))$
$\begin{array} ee\equiv(a\lor d)\\ f\equiv(\lnot a)\\ g\equiv(f\land b)\\ h\equiv(\lnot c)\\ i\equiv(h\land d)\\ j\equiv(g\lor i)\\ k\equiv(e\land j)\\ l\equiv(a\land c)\\ m\equiv(b\land d)\\ n\equiv(l\lor m)\\ o\equiv(\lnot n)\\ \hline F\equiv(k\land o) \end{array}$
$\begin{array}{cccc|ccccccccccc|c} a&b&c&d&e&f&g&h&i&j&k&l&m&n&o&F\\ \hline 0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0\\ 0&0&0&1&1&1&0&1&1&1&1&0&0&0&1&1\\ 0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0\\ 0&0&1&1&1&1&0&0&0&0&0&0&0&0&1&0\\ 0&1&0&0&0&1&1&1&0&1&0&0&0&0&1&0\\ 0&1&0&1&1&1&1&1&1&1&1&0&1&1&0&0\\ 0&1&1&0&0&1&1&0&0&1&0&0&0&0&1&0\\ 0&1&1&1&1&1&1&0&0&1&1&0&1&1&0&0\\ 1&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\ 1&0&0&1&1&0&0&1&1&1&1&0&0&0&1&1\\ 1&0&1&0&1&0&0&0&0&0&0&1&0&1&0&0\\ 1&0&1&1&1&0&0&0&0&0&0&1&0&1&0&0\\ 1&1&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\ 1&1&0&1&1&0&0&1&1&1&1&0&1&1&0&0\\ 1&1&1&0&1&0&0&0&0&0&0&1&0&1&0&0\\ 1&1&1&1&1&0&0&0&0&0&0&1&1&1&0&0 \end{array}$
This is a step-by-step truth table of your logical function. As you can see, somewhere in your parsing you came up with an extra term.
Senex Ægypti Parvi
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